no suitable constructor exists to convert from "test *" to "test", constructor,

Question

I'm new to c++ and I'm having difficulties with constructor and classes. So, here is my header file:

#pragma once
#include <string>
using namespace std;
class test
{

    private:
    string name;
    int number;

public:

    test();
    test(string i,int b);
};

This is cpp file:

#include "test.h"
#include <string>
using namespace std;


test::test(){}

test::test(string i,int b){
    this->name=i;
    this->number=b;
}

now, when I try to call

test t=new test("rrr",8);

I get:

1   IntelliSense: no suitable constructor exists to convert from "test *" to "test" 

So, whats the thing with classes having * in their name ( for instance, classes without .cpp file don't have asterix, all others do)? And what do I do wrong?

Answer 1

I imagine you're coming from a Java/C# background. t is not a reference type here, it's a value type. new returns a pointer to an object. So you need any of the following:

test t = test("rrr", 8);
test t("rrr", 8);
test *t = new test("rrr", 8);

If you're not yet familiar with pointers, then definitely don't use the last one! But understanding the semantics of pointers is fairly critical; I recommend reading the relevant chapter(s) in your textbook...

Answer 2

So, whats the thing with classes having "*" in their name ( for instance, classes without .cpp file dont have asterix, all others do)???

You definitely need to learn about pointers. test * and test are two completely different types in C++. Here's two variables with those types:

test t;
test* p;

Here, t has type test, and p as type test*. We describe test* as "pointer to test".

You can often think of a pointer as being the memory address of an object. So in p, since it is a pointer, we could store the memory address of t, which is a test. To get the address of an object, we use the unary & operator, like so:

test t;
test* p = &t;

Note that t is a test object. You didn't need to say new test(). This is where C++ differs from other languages that you might have used, like C# and Java. In the above C++ code, t is a test object.

However, you can create objects with new test(), so what's the difference?

test t; creates a test object with automatic storage duration. This means it is destroyed at the end of its scope (often the function is being declared within).

new test() creates a test object with dynamic storage duration. This means you have to destroy the object manually, otherwise you'll have a memory leak. This expression returns a pointer and so you can initialise a pointer object with it:

test* p = new test();

So now let's look at your problem:

test t=new test("rrr",8);

We now know that new test("rrr", 8) returns a pointer to test (a test*). However, you're trying to assign it to a test object. You simply can't do this. One of them is an address and the other is a test. Hence the compiler says "no suitable constructor exists to convert from test * to test." Makes sense now, doesn't it?

Instead, you should prefer to use automatic storage duration. Only use new if you really really need to. So just do:

test t("rrr", 8);

Answer 3

test t=new test("rrr",8);

must be

//  v
test* t=new test("rrr",8);

---

So, whats the thing with classes having "*" in their name

* is used to indicate a pointer, it's not in the name of the class. But it's a big topic, so you should do some reseach on this.